Difference between revisions of "1992 OIM Problems/Problem 2"
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[[File:1992_OIM_P2a.png|center|600px]] | [[File:1992_OIM_P2a.png|center|600px]] | ||
− | Notice that the intervals will be: <math>I_1=r_1-(-a_1), I_2=r_2-(- | + | Notice that the intervals will be: <math>I_1=r_1-(-a_1), I_2=r_2-(-a_2), \cdots , I_n=r_n-(-a_n)</math> |
Thus the sum of the intervals will be: <math>\sum_{i}^{}\left( r_i+a_i \right)</math> | Thus the sum of the intervals will be: <math>\sum_{i}^{}\left( r_i+a_i \right)</math> |
Revision as of 12:02, 17 December 2023
Problem
Given the collection of positive real numbers
and the function:
Determine the sum of the lengths of the intervals, disjoint two by two, formed by all .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Since , we can plot
to visualize what we're looking for:
Notice that the intervals will be:
Thus the sum of the intervals will be:
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I got a ZERO on this one because I didn't even know what was I supposed to do, nor did I know what the sum of the lengths of the intervals, disjoint two by two meant. A decade ago I finally solved it but now I don't remember how. I will attempt to solve this one later.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.